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First-order differential equation

  1. May 22, 2005 #1
    A ball with mass [tex]0.15\mbox{ kg}[/tex] is thrown upward with initial velocity [tex]20\mbox{ m/s}[/tex] from the roof of a building [tex]30 \mbox{ m}[/tex] high. There is a force due to air resistance of [tex]\frac{v^2}{1325}[/tex], where the velocity [tex]v[/tex] is measured in [tex]\mbox{ m/s}[/tex].

    I've deliberately disregarded the rest of the problem. My question is about the first-order differential equation I set up based upon the given information.

    [tex]m\frac{dv}{dt}=-mg-\frac{v^2}{1325}, \qquad v(0)=v_0 \quad x(0)=x_0[/tex]

    [tex]\frac{dv}{dt} + \frac{v^2}{1325m} = -g[/tex]

    As far as I can see, the method of integrating factors does not work here. I don't know what to do. It is possible that I made a mistake in the D.E. set up.

    Any help is highly appreciated.
     
  2. jcsd
  3. May 22, 2005 #2

    dextercioby

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    HINT:Variables are separable.

    Daniel.
     
  4. May 22, 2005 #3
    Oh, I see. Thanks.

    Here's what I have now:

    [tex]\frac{dv}{dt}=-\frac{v^2}{1325m}-g[/tex]

    [tex]\frac{dv}{dt}=-\frac{\left( v^2 + 1325 mg \right)}{1325m}[/tex]

    [tex]\int \frac{dv}{v^2 + 1325 mg} = -\frac{1}{1325m}\int dt[/tex]

    On the left-hand side, the substitution [tex]v=\sqrt{1325 mg}\tan \theta[/tex] gives

    [tex]\int \frac{dv}{v^2 + 1325 mg} = \frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v \right) + \mathrm{C}[/tex]

    and so we have

    [tex]\frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v \right) = -\frac{t}{1325m} + \mathrm{C}[/tex]

    [tex]v(0)=v_0 \Rightarrow \mathrm{C} = \frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right)[/tex]
     
    Last edited: May 22, 2005
  5. May 22, 2005 #4

    dextercioby

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    Yes,it looks okay.

    Daniel.
     
  6. May 22, 2005 #5

    arildno

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    Keep in mind that this solution only holds on the way up...
     
    Last edited: May 22, 2005
  7. May 22, 2005 #6
    I've got some more, but I'm not sure it is correct.

    [tex]\frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v \right) = \frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right) -\frac{t}{1325m}[/tex]

    [tex]\arctan \left( \frac{\sqrt{1325mg}}{1325mg}v \right) = \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right) -\frac{gt}{\sqrt{1325mg}} [/tex]

    [tex]v(t) = \sqrt{1325mg} \tan \left[ \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right) -\frac{gt}{\sqrt{1325mg}} \right][/tex]

    [tex]x(t) = \sqrt{1325mg} \int \tan \left[ \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right) -\frac{gt}{\sqrt{1325mg}} \right] \: dt[/tex]

    Consider the integral

    [tex]\int \tan (a-bt) \: dt[/tex]

    The substitution [tex]u=a-bt[/tex] gives

    [tex]\int \tan (a-bt) \: dt = - \frac{1}{b} \int \tan u \: du = - \frac{1}{b} \ln \left| \sec u \right| + \mathrm{C} = - \frac{1}{b} \ln \left| \sec (a-bt) \right| + \mathrm{C}[/tex]

    Thus, we have

    [tex]x(t)= 1325m \ln \left| \sec \left[ \arctan \left( \frac{\sqrt{1325mg}}{1325mg} v_0 \right) -\frac{gt}{\sqrt{1325mg}} \right] \right| + \mathrm{C}[/tex]

    [tex]x(0)=x_0 \Rightarrow \mathrm{C} = x_0 + \frac{1325m}{2}\ln \left( \frac{1325mg}{1325mg + v_0 ^2} \right)[/tex]

    [tex]x(t)= 1325m \ln \left| \sec \left[ \arctan \left( \frac{\sqrt{1325mg}}{1325mg} v_0 \right) -\frac{gt}{\sqrt{1325mg}} \right] \right| + x_0 + \frac{1325m}{2}\ln \left( \frac{1325mg}{1325mg + v_0 ^2} \right)[/tex]

    Thanks
     
  8. May 22, 2005 #7

    arildno

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    Remember that [tex]\cos(arctan(y))=\frac{1}{\sqrt{1+{y}^{2}}}[/tex]

    This will give you [tex]x(0)=x_{0}[/tex] as it should..
     
  9. May 23, 2005 #8
    That's what I did up there! Unfortunately, when I use the [tex]v(t)[/tex] and [tex]x(t)[/tex] above, I get some wierd results.

    Let's say I wanna find the maximum height above the ground the ball reaches. That happens when [tex]\frac{dx}{dt}=v(t)=0[/tex]. I simply need to solve for [tex]t[/tex] and plug it into [tex]x(t)[/tex]. I get [tex]t \approx 1.9161467809752109 \mbox{ s}[/tex] and [tex]x(t) \approx 11.438482854170548 \mbox{ m}[/tex]. The answer in my textbook is [tex]x(t) \approx 48.562 \mbox{ m}[/tex].

    Let's now suppose I want to find the time when the ball hits the ground. I set [tex]x(t)=0[/tex] and solve for [tex]t[/tex]. The problem now is that I get complex numbers, while the answer in my textbook is [tex]t \approx 5.194 \mbox{ s}[/tex]

    There must be some mistake when I try to integrate v(t), but I don't know exactly where it is.

    Thanks
     
  10. May 23, 2005 #9
    First off, let me rectify my previous post. When I use the [tex]x(t)[/tex] and [tex]v(t)[/tex] above to find:

    (a) the maximum height above the ground the ball reaches, I should have instead

    [tex]v(t)=0[/tex]. Then, solving for [tex]t[/tex] gives

    [tex]t\approx 1.9161467809752113 \mbox{ s}[/tex]

    and so

    [tex]x(t) \approx 48.561517145829434 \mbox{ m}[/tex]

    (b) the time when the ball hits the ground, I should have instead

    [tex]x(t)=0[/tex]. Then, solving for [tex]t[/tex] gives

    [tex]t\approx 4.937745892290328 \mbox{ s}[/tex]

    The good news is that I get the exact same answers with the [tex]x(t)[/tex] and [tex]v(t)[/tex] I obtain with the aid of Mathematica's command DSolve, which solves differential equations. I just needed to enter this:

    [tex]m\frac{dv}{dt}=-mg-\frac{v^2}{1325}, \qquad v(0)=v_0 \quad x(0)=x_0[/tex]

    which is the same I used to solve it myself. Thus, the mistake must either be in the differential equation set up I did or in the answers provided by the textbook. Let me reiterate what the textbook gives

    (a) [tex]x(t) \approx 48.562 \mbox{ m}[/tex]

    (b) [tex]t \approx 5.194 \mbox{ s}[/tex]

    I still think it is much more likely that I made the mistake. Can anybody please help me figure this out?

    Thanks
     
    Last edited: May 23, 2005
  11. May 23, 2005 #10

    arildno

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    You got a) right then after all!

    As for b), did you read my post 5?

    Remember that air resistance is always in the direction OPPOSITE to the velocity the object has through the air!

    Thus, the air resistance acting upon the object on the way down cannot be [tex]-kv^{2}[/tex] where k is some constant gravity acts along the negative axis.

    The correct quadratic air resitance law is: [tex]-k|v|v[/tex], where |v| is the speed of the object.
    Thus, on its way down, Newton's second law reads in your case (remember v<0 now!):
    [tex]m\frac{dv}{dt}=-mg+\frac{v^{2}}{1325}[/tex]
    Got it?

    As a check, you'll end up with an expression involving the inverse of the hyperbolic tangent function, rather than arctan.
     
    Last edited: May 23, 2005
  12. May 24, 2005 #11
    Now it makes sense! Thank you

    [tex]m\frac{dv}{dt}=-mg+\frac{v^2}{1325}[/tex]

    [tex]\frac{dv}{dt}=-g+\frac{v^2}{1325m}[/tex]

    [tex]\frac{dv}{dt}=\frac{v^2 - 1325mg}{1325m}[/tex]

    [tex]\int \frac{dv}{v^2 - 1325mg} = \frac{1}{1325m}\int dt[/tex]

    Consider the left-hand side:

    [tex]\int \frac{dv}{v^2 - 1325mg}=-\frac{1}{1325mg} \int \frac{1325mg}{1325mg-v^2} \: dv[/tex]

    [tex]\int \frac{dv}{v^2 - 1325mg}=-\frac{1}{1325mg} \int \frac{dv}{1-\left( \frac{v}{\sqrt{1325mg}} \right) ^2}[/tex]

    [tex]\int \frac{dv}{v^2 - 1325mg}=-\frac{1}{1325mg} \mbox{arctanh} \left( \frac{v}{\sqrt{1325mg}} \right) + \mathrm{C}[/tex]

    Thus, we have

    [tex]-\frac{1}{1325mg} \mbox{arctanh} \left( \frac{v}{\sqrt{1325mg}} \right) = \frac{t}{1325m} + \mathrm{C}[/tex]

    [tex]\mbox{arctanh} \left( \frac{v}{\sqrt{1325mg}} \right) = -\frac{gt}{\sqrt{1325mg}} + \mathrm{C}[/tex]

    [tex]v(t) = \sqrt{1325mg} \tanh \left( -\frac{gt}{\sqrt{1325mg}} + \mathrm{C} \right)[/tex]

    [tex]v(0)=0 \Rightarrow \mathrm{C}=0[/tex]

    [tex]v(t) = \sqrt{1325mg} \tanh \left( -\frac{gt}{\sqrt{1325mg}} \right)[/tex]

    and so

    [tex]x(t) = \sqrt{1325mg} \int \tanh \left( -\frac{gt}{\sqrt{1325mg}} \right) \: dt[/tex]

    The substitution

    [tex]u=-\frac{gt}{\sqrt{1325mg}}[/tex]

    gives

    [tex]x(t) = -1325m \int \tanh u \: du[/tex]

    [tex]x(t) = -1325m \ln (\cosh u) + \mathrm{C}[/tex]

    [tex]x(0)=x_{\mbox{max}} \Rightarrow \mathrm{C} = x_{\mbox{max}}[/tex]

    [tex]x(t) = -1325m \ln \left[ \cosh \left( \frac{gt}{\sqrt{1325mg}} \right) \right] + x_{\mbox{max}}[/tex]

    Now, we can find the time for the ball to fall from the maximum height

    [tex]x(t)=0 \Rightarrow t \approx 3.277723004542093 \mbox{ s}[/tex]

    which is added to the time found in part (a)

    [tex]t \approx 1.9161467809752113 \mbox{ s}[/tex]

    and gives the total time

    [tex]t \approx 5.194 \mbox{ s}[/tex]
     
    Last edited: May 24, 2005
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